Determine \( f_{x} \) und \( f y \) of the given function \( f(x, y)=\left(x^{5} \ln 2 x\right) y^{8} \)

To find the partial derivatives \( f_x \) and \( f_y \) of the function \[ f(x, y) = \left(x^{5} \ln(2x)\right) y^{8}, \] we will differentiate with respect to \( x \) and \( y \) separately, treating the other variable as a constant in each case. ### Partial Derivative with Respect to \( x \) (\( f_x \)) \[ f(x, y) = x^{5} \ln(2x) \cdot y^{8} \] To differentiate \( f \) with respect to \( x \), treat \( y^8 \) as a constant: \[ f_x = y^{8} \cdot \frac{d}{dx}\left(x^{5} \ln(2x)\right) \] Use the **product rule** for differentiation, where if \( u = x^5 \) and \( v = \ln(2x) \), then: \[ \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \] Compute \( u' \) and \( v' \): \[ u' = \frac{d}{dx}(x^5) = 5x^4 \] \[ v' = \frac{d}{dx}(\ln(2x)) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \] Applying the product rule: \[ \frac{d}{dx}\left(x^{5} \ln(2x)\right) = 5x^{4} \ln(2x) + x^{5} \cdot \frac{1}{x} = 5x^{4} \ln(2x) + x^{4} \] Factor out \( x^4 \): \[ f_x = y^{8} \left(5x^{4} \ln(2x) + x^{4}\right) = y^{8} x^{4} \left(5 \ln(2x) + 1\right) \] ### Partial Derivative with Respect to \( y \) (\( f_y \)) \[ f(x, y) = x^{5} \ln(2x) \cdot y^{8} \] To differentiate \( f \) with respect to \( y \), treat \( x^5 \ln(2x) \) as a constant: \[ f_y = x^{5} \ln(2x) \cdot \frac{d}{dx}(y^{8}) = x^{5} \ln(2x) \cdot 8y^{7} = 8x^{5} \ln(2x) y^{7} \] ### Summary The partial derivatives of \( f(x, y) = \left(x^{5} \ln(2x)\right) y^{8} \) are: \[ f_x = y^{8} x^{4} \left(5 \ln(2x) + 1\right) \] \[ f_y = 8x^{5} \ln(2x) y^{7} \]